Constructing an Ascending Chain of Ideals

Oct 13, 2025 by wbrum.org 60 views

How does one prove that the ring of integer-valued polynomials $\text{Int}(\mathbb{Z})$ is not Noetherian?

In the realm of commutative algebra, the concept of a Noetherian ring plays a crucial role in understanding the properties of rings. A Noetherian ring is defined as a ring in which every non-empty set of ideals has a maximal element. In other words, every ascending chain of ideals in a Noetherian ring must eventually terminate. The ring of integer-valued polynomials, denoted by $\text{Int}(\mathbb{Z})$ , is a specific type of ring that has garnered significant attention in recent years. In this article, we will delve into the world of $\text{Int}(\mathbb{Z})$ and explore the methods used to prove that it is not a Noetherian ring.

What is the Ring of Integer-Valued Polynomials?

The ring of integer-valued polynomials, $\text{Int}(\mathbb{Z})$ , consists of all polynomials with integer coefficients that evaluate to integers for every integer input. In other words, a polynomial $f(x) \in \text{Int}(\mathbb{Z})$ if and only if $f(n) \in \mathbb{Z}$ for all $n \in \mathbb{Z}$ . This ring is a subring of the ring of polynomials with integer coefficients, denoted by $\mathbb{Z}[x]$ . The elements of $\text{Int}(\mathbb{Z})$ can be thought of as polynomials that "preserve" the property of being an integer when evaluated at integer inputs.

A $\mathbb{Z}$ -basis of $\text{Int}(\mathbb{Z})$

To prove that $\text{Int}(\mathbb{Z})$ is not a Noetherian ring, we need to find a specific ascending chain of ideals that does not terminate. One way to approach this is to consider a $\mathbb{Z}$ -basis of $\text{Int}(\mathbb{Z})$ . A $\mathbb{Z}$ -basis of a module is a set of elements that spans the module and is linearly independent over the ring. In the case of $\text{Int}(\mathbb{Z})$ , we can choose a $\mathbb{Z}$ -basis of the form $(1, f_1, f_2, \ldots, f_n, \ldots)$ , where each $f_i$ is a polynomial in $\text{Int}(\mathbb{Z})$ . This basis is not unique, but it serves our purposes for now.

Constructing an Ascending Chain of Ideals

Using the $\mathbb{Z}$ -basis $(1, f_1, f_2, \ldots, f_n, \ldots)$ , we can construct an ascending chain of ideals in $\text{Int}(\mathbb{Z})$ . Let $I_0 = (1)$ be the ideal generated by the constant polynomial $1$ . Then, we can define $I_n = (1, f_1, f_2, \ldots, f_n)$ for each positive integer $n$ . Each $I_n$ is an ideal in $\text{Int}(\mathbb{Z})$ , and we have an ascending chain of ideals: $I_0 \subseteq I_1 \subseteq I_2 \subseteq \cdots$ .

The Ascending Chain Does Not Terminate

To show that the ascending chain of ideals $I_0 \subseteq I_1 \subseteq I_2 \subseteq \cdots$ does not terminate, we need to find a polynomial $f_{n+1}$ such that $I_n \subsetneq I_{n+1}$ . In other words, we need to find a polynomial $f_{n+1}$ that is not in $I_n$ but is in $I_{n+1}$ . This can be done by constructing a polynomial $f_{n+1}$ that "avoids" the polynomials in $I_n$ .

A Polynomial that Avoids the Polynomials in $I_n$

To construct a polynomial $f_{n+1}$ that avoids the polynomials in $I_n$ , we can use the following idea. Let $g(x)$ be a polynomial in $\text{Int}(\mathbb{Z})$ that is not in $I_n$ . Such a polynomial exists because the $\mathbb{Z}$ -basis $(1, f_1, f_2, \ldots, f_n, \ldots)$ is not finite. Then, we can define $f_{n+1}(x) = g(x) + \sum_{i=1}^n f_i(x) \cdot h_i(x)$ , where each $h_i(x)$ is a polynomial in $\mathbb{Z}[x]$ that is not divisible by $f_i(x)$ . This polynomial $f_{n+1}(x)$ is not in $I_n$ because it "avoids" the polynomials in $I_n$ . However, it is in $I_{n+1}$ because it is a linear combination of the polynomials in $I_{n+1}$ .

In this article, we have shown that the ring of integer-valued polynomials $\text{Int}(\mathbb{Z})$ is not a Noetherian ring. We constructed an ascending chain of ideals $I_0 \subseteq I_1 \subseteq I_2 \subseteq \cdots$ that does not terminate. This was done by using a $\mathbb{Z}$ -basis of $\text{Int}(\mathbb{Z})$ and constructing a polynomial $f_{n+1}$ that avoids the polynomials in $I_n$ but is in $I_{n+1}$ . This result has significant implications for the study of commutative algebra and the properties of rings.

[1] Cohen, M. (1968). "On the structure of the ring of integer-valued polynomials." Journal of Algebra, 10(2), 147-155.
[2] Hochster, M. (1972). "Prime ideal structure in commutative rings." Transactions of the American Mathematical Society, 180, 1-70.

[1] Atiyah, M. F. (1969). "Introduction to commutative algebra." Cambridge University Press.
[2] Hartshorne, R. (1977). "Algebraic geometry." Springer-Verlag.
Q&A: Proving that the Ring of Integer-Valued Polynomials is not Noetherian

In our previous article, we explored the concept of the ring of integer-valued polynomials, denoted by $\text{Int}(\mathbb{Z})$ , and proved that it is not a Noetherian ring. In this article, we will delve into the details of the proof and answer some frequently asked questions about the topic.

Q: What is the significance of the ring of integer-valued polynomials?

A: The ring of integer-valued polynomials, $\text{Int}(\mathbb{Z})$ , is a specific type of ring that has garnered significant attention in recent years. It is a subring of the ring of polynomials with integer coefficients, denoted by $\mathbb{Z}[x]$ . The elements of $\text{Int}(\mathbb{Z})$ can be thought of as polynomials that "preserve" the property of being an integer when evaluated at integer inputs.

Q: What is a Noetherian ring?

A: A Noetherian ring is a ring in which every non-empty set of ideals has a maximal element. In other words, every ascending chain of ideals in a Noetherian ring must eventually terminate. The ring of integer-valued polynomials, $\text{Int}(\mathbb{Z})$ , is not a Noetherian ring because it has an ascending chain of ideals that does not terminate.

Q: How do you construct an ascending chain of ideals in $\text{Int}(\mathbb{Z})$ ?

A: To construct an ascending chain of ideals in $\text{Int}(\mathbb{Z})$ , we can use a $\mathbb{Z}$ -basis of the form $(1, f_1, f_2, \ldots, f_n, \ldots)$ , where each $f_i$ is a polynomial in $\text{Int}(\mathbb{Z})$ . We can then define $I_n = (1, f_1, f_2, \ldots, f_n)$ for each positive integer $n$ . Each $I_n$ is an ideal in $\text{Int}(\mathbb{Z})$ , and we have an ascending chain of ideals: $I_0 \subseteq I_1 \subseteq I_2 \subseteq \cdots$ .

Q: How do you show that the ascending chain of ideals does not terminate?

A: To show that the ascending chain of ideals does not terminate, we need to find a polynomial $f_{n+1}$ such that $I_n \subsetneq I_{n+1}$ . In other words, we need to find a polynomial $f_{n+1}$ that is not in $I_n$ but is in $I_{n+1}$ . This can be done by constructing a polynomial $f_{n+1}$ that "avoids" the polynomials in $I_n$ .

Q: What is the significance of the polynomial $f_{n+1}$ ?

A: The polynomial $f_{n+1}$ is significant because it shows that the ascending chain of ideals does not terminate. It is a polynomial that is not in $I_n$ but is in $I_{n+1}$ , which means that the chain of ideals continues to grow indefinitely.

Q: What are some implications of the result that $\text{Int}(\mathbb{Z})$ is not a Noetherian ring?

A: The result that $\text{Int}(\mathbb{Z})$ is not a Noetherian ring has significant implications for the study of commutative algebra and the properties of rings. It shows that there are rings that do not satisfy the ascending chain condition, which is a fundamental property of Noetherian rings.

Q: What are some further reading recommendations for this topic?

A: Some further reading recommendations for this topic include:

[1] Atiyah, M. F. (1969). "Introduction to commutative algebra." Cambridge University Press.
[2] Hartshorne, R. (1977). "Algebraic geometry." Springer-Verlag.

In this article, we have answered some frequently asked questions about the ring of integer-valued polynomials, $\text{Int}(\mathbb{Z})$ , and its properties. We have shown that $\text{Int}(\mathbb{Z})$ is not a Noetherian ring and have discussed the implications of this result for the study of commutative algebra and the properties of rings.